Codeforces Round #246 (Div. 2) B. Football Kit

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B. Football Kit

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Consider a football tournament where n teams participate. Each team has two football kits: for home games, and for away games. The kit for home games of the i-th team has color xi and the kit for away games of this team has color yi (xi ≠ yi).

In the tournament, each team plays exactly one home game and exactly one away game with each other team (n(n - 1) games in total). The team, that plays the home game, traditionally plays in its home kit. The team that plays an away game plays in its away kit. However, if two teams has the kits of the same color, they cannot be distinguished. In this case the away team plays in its home kit.

Calculate how many games in the described tournament each team plays in its home kit and how many games it plays in its away kit.

Input

The first line contains a single integer n (2 ≤ n ≤ 105) — the number of teams. Next n lines contain the description of the teams. Thei-th line contains two space-separated numbers xi, yi (1 ≤ xi, yi ≤ 105; xi ≠ yi) — the color numbers for the home and away kits of thei-th team.

Output

For each team, print on a single line two space-separated integers — the number of games this team is going to play in home and away kits, correspondingly. Print the answers for the teams in the order they appeared in the input.

Sample test(s)

input

21 22 1

output

2 02 0

input

31 22 11 3

output

3 14 02 2

题意太绕，我看了半天才看懂，太难懂啊太难懂

大致是：每个队伍可以参加主场比赛（n-1）场和客场比赛（n-1）场，

每个队伍打主场比赛的时候穿主场衣服，打客场比赛的时候穿客场衣服，

但是当客场衣服与主场衣服相同的时候，客场穿主场衣服，

求每个队的(穿主场衣服的次数)和(穿客场衣服的次数)

所以 主场衣服次数=（n-1）+客场衣服与其他主场衣服相同的次数

         客场衣服次数=2*（n-1）-主场衣服次数

                                =（n-1）-客场衣服与其他主场衣服相同的次数

代码:

#include<iostream>#include<algorithm>#include<queue>#include<stack>#include<cmath>#include<string.h>#include<stdio.h>#include<stdlib.h>using namespace std;int x[100005],y[100005];int a[100005];int main(){    int n,i,j,k;    while(~scanf("%d",&n))    {        memset(a,0,sizeof(a));      for(i=0;i<n;i++)      {          scanf("%d%d",&x[i],&y[i]);          a[x[i]]++;      }      for(i=0;i<n;i++)      {          int m=n-1+a[y[i]];          int t=n-1-a[y[i]];          printf("%d %d\n",m,t);      }    }    return 0;}